$W$ is not equal to $W^{\perp\perp}$?

Asked Nov 10 '16 at 15:51

Active Nov 10 '16 at 15:57

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Let $V$ be a vector space with inner product. If $W$ is a closed subspace of $V$ we always have $W\subseteq W^{\perp\perp}$. Now, I found a proof where the equality holds if $V$ is Hilbert. It is an easy consequence of $V=W\oplus W^{\perp}$. So my question is if this is not true if $V$ is not Hilbert. Does anyone know a counterexample?

Thank you.

Edit: Oh, sorry if I didn't clarify. I'm still suppossing $W$ is closed.

edited Nov 10 '16 at 15:57

asked Nov 10 '16 at 15:51

Talexius

2,105

1

In general $W^{\perp\perp}=\overline{W}$. Try your favourite Banach space that is not a Hilbert space, maybe you can find an example. – Mathematician 42 Nov 10 '16 at 15:53
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@Mathematician42 but it's given that $W$ is closed – Ben Grossmann Nov 10 '16 at 15:54

$W$ is not equal to $W^{\perp\perp}$?

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